You can actually calculate the rate at which a wing in steady flight changes the downward momentum of "the air". In order to perform this calculation though you need to be very specific about what you mean by "the air". If you decide to consider consider "all" of the air affected by the wing (a sensible approach), you can do this by pushing the boundaries of the "control volume" you use in your analysis out to infinity. This approach gives a somewhat surprising result. Even if the closest boundary of your control volume is an infinite number of spans from the wing, the calculated rate of chage of the air's downward momentum depends on the SHAPE of the control volume. If the front and back (upstream and downstream) faces of the control volume are very tall and skinny, the rate of change of the air's downward momentum is equal to the lift (the pressure footprints on the top/bottom of the control volume contribute no net vertical force in this case). If the front and back faces of the control volume are very short and fat, the rate of change of the air's downward momentum is equal to zero (the pressure footprints on the top/bottom of the control volume contribute a net vertical force equal to the lift in this case). There's no paradox here... the results of either case are entirely consistent with Newton's Laws. In the tall/skinny case the air experiences an unbalanced force from the wing and a corresponding rate of momentum change. In the short/fat case the force exerted by the wing on the air is balanced by an equal and opposite net force exerted by the top/bottom of the control volume.

The idea that a wing's lift is balanced by the rate of downward momentum transfer to the downwash is very appealing (to me anyway). Unfortunately, the reality ain't quite that simple.

(10) Are you proposing that the rudder does not work because it has no earth to push on?

Tom

No, you are confused about my claims of cause and effect, or I have not stated them clearly. I never said that that the wing works only because the earth pushes back. A wing would still generate lift as a spacecraft passed through a cloud of molecules with no earth nearby. The downwash in this case would continue forever-- the vortices would detach from the gas cloud and keep going forever. My original point all along with this thread, is that if the earth does not somehow feel the downward "push" of the plane in some way or the other-- and I am now realizing that that way of feeling the plane's downward push need not involve momentum-- then the plane's gravity will attract the earth upward and momentum will not be conserved. See posts #22 and #25. I still stand by that claim 100%.

As far as point #3 goes, surely we all agree that if air is exerting an upforce on the wing, then the wing is exerting a downforce on the air? I'm sure we all do.

The whole point of yesterday's post is simply to say that the downforce exerted by the wing onto the air is resisted by the air's resistance to shearing, and so opposing forces arise that greatly reduce the NET downward force on the air, so the air's NET downward acceleration is much less than would be the case without the opposing shearing forces. If we were simply creating propulsive force by, say, shooting bb's out of cannon into the vacuum of space, then the rearward momentum imparted to the bb's is the same as the forward momentum imparted to the spacecraft. But when we complicate the picture with a surrounding fluid, opposing forces arise and the picture becomes more complex. For example consider propwash. The air is exerting a forward force on the aircaft, but in steady cruise this force does not accelerate the aircraft forward, because it is opposed by other forces (drag) and the net forward force on the aircraft is zero. Likewise the prop exerts a backwards force on the air, but other opposing forces are exerted by the fluid on the air molecules near the prop, so the net acceleration of the air near the prop is less than would be predicted in the absence of these opposing forces. To take an extreme case, if we were flying in honey or molasses or jello, then the prop would "screw" its way through the fluid with very little slippage and very little rearward momentum would be imparted by the prop to the fluid. Likewise, if we are flying through honey, the honey's resistance to shearing greatly reduces the downwash velocity (and the downwash momentum) imparted by the wing to the fluid. Yet the wing is still exerting a downforce on the honey, which is still "felt" by a scale sitting underneath the vat of honey.

Just as when we stand on solid ground, our feet exert a downforce on the ground, just as the ground exerts an upforce on our feet, but there is no momentum transfer from our feet to the ground, because the ground is very viscous, you might say.

You're still tied to the idea that the plane has to push against the earth. This is simply not so. The air that the airplane flies through has mass of its own. That mass is what reacts to the airplane. The air is not simply a conductive medium connecting the airplane to the ground.

If the airplane needed something to "push against" other than air in order to complete this huge circle of effect then pulling a positive G load going over the top of a high speed loop would not be possible. Nor would pulling positive G when in the vertical portions of the loop. And nor would Tom's rudder. Yet these things all work even when there's no "earth" to push against in the direction that the lift is being made at that time. The air has mass and it's enough that the plane can push and pull against that mass.

The earth's only role in a plane flying level is that it produces a constant downward attractive force of 1G. And that means that the airplane has to constantly develop 1G of lift to counteract the force of gravity.

So where does the force generated by the plane go once it's "skipped" off the air and left behind some downwash? That's the simple one. The answer is simply "heat". Air is viscous and thus it tends to damp out any kinetic motion it has. This damping out of the motion within the body of air is why you can stand under the glide path of a big jet that is passing by at a mere couple of hundred feet up and not feel the downwash on the ground.

You're also assuming that there is no other effect than the downwash left in the wake of a plane. If you consider that a wing and mass of a plane passing through a volume of air produces a downwash that implies that the air moving down and away from where it was before has to produce a matching low pressure area in the body of air that the mass is moving FROM. This lower pressure above the downwash is going to also try to slow down and damp out the downward momentum. And in fact if you look at air coming off the rear of airfoils in a smoke trail wind tunnel it's pretty apparent that any downwash is damped out and slowed down to a great extent by the time the air has moved behind the wing by as little as one chord length.

Movies of aircraft flying in close proximity to clouds further supports this idea that the plane's lift is a product of interaction with the air alone. A plane has to be darn close to a cloud to produce a disturbance in the formation. By the time it's two to three wingspans away there's no effect. So again it shows that the airmass itself is able to deal with the needs for flying the plane without the need to connect with the earth.

For the same reason your space craft example would not produce a downwash velecity vector which traveled for thousands of miles or light years.

I created this airfoil in XFLR5 and analyzed it at an AOA of 7 degrees. The results show a Cl of +0.122 (so the force on the airfoil is upward), but if you look at the boundary layer, you can plainly see the air is being deflected upward behind the airfoil. We all know that an airfoil can't be lifting unless it is deflecting air downward, so when are they going to fix XFLR5?

The reflexed profiles used in flying wings rely on low pressure on the lower portion of the trailing edge to counteract the wing pitching moment. this means that the air leaving the trailing edge is deflected upward, though the frontal portion of the wing will act like any other section. Essentially the wing will draw the freestream upward, bend it downward as it goes over the wing, then give it a small kick upward to avoid pitching down. Overall, the air that gets disturbed stays on average in the same place, like for a normal wing section. you can play with the system boundaries to get the force vectors on the wing basing yourself only on reaction, or basing yourself only on pressure differences... the interaction between all these things is complex, there is no one single reason why lift is "made". Lift is the result of an interplay of several things all influencing each other in a complex system, and you can only really measure how these items interoperate, and try to predict their behaviour.

You're still tied to the idea that the plane has to push against the earth. This is simply not so.

Define the meaning of "has to". I don't mean that lift is tied to the creation of a push against the earth-- for example, a rudder creates sideways force without any push against the earth. But, the rudder can't make a net sideforce indefinitely without making the flight path curve. Curving flight is not a valid inertial reference frame.

What I mean by "has to", is that if the plane did not push against the earth, the plane's gravitational pull would accelerate the earth upwards. This obviously violates the laws of conservation of momentum. For example, in a few thousand years, the plane would accelerate the earth to a delta v comparable with the earth's orbital velocity around the sun. This is nontrivial. It doesn't make sense that this should happen. It doesn't happen. Because the earth does "feel" the aircraft's weight.

To be perfectly consistent, assume an earth far from any sun. Not describing a curved path through space. Only travelling in a straight line. This earth is a valid inertial reference frame. A plane creating lift over this earth would accelerate this earth upwards, by gravitational attraction, if the earth did not "feel" a downward push under the plane equal in magnitude to the plane's weight. This would violate the law of conservation of momentum.

This cannot be. So, we are forced to conclude that the earth does feel a downward push equal in magnitude to the upward gravitational attraction exerted by the plane, i.e. equal in magnitude to the plane's weight.

The reason the earth "feels" the wing's downforce but does not "feel" the rudder's sideforce, is that the rudder's sideforce is not aimed at the earth, while the wing's downforce is aimed squarely at the earth, except during transitory manuevers such as turns and loops. During a turn, a 1-G downforce component is still aimed at the earth. During a loop the average downforce component is still 1-G. So over the long run, the earth "feels" a 1-G downforce (equal to the weight of the plane) even during these maneuvers.

Surely this is obvious. If we have a lightweight framework enclosing a cube of air one mile per side, open on top, and a plane is flying in the cube of air, a scale under the cube of air measures more weight when the plane is engaged in a 2-G pullout, then when the plane is engaged in a zero-G pushover. Does anyone doubt that? It seems obvious to me.

Over the long run, on average, the scale "feels" a force equal to the weight of the plane.

Just as a scale under an athlete registers more weight when the athlete's legs are accelerating him/ her upward at 2G's, than when his/her leg muscles go slack and he/she accelerates earthwards with no resistance to gravity (0-G).

But over the long run, the scale registers a force equal to the athlete's weight.

I realize that this rocks some people's paradigms and they are forced as a result to use the "waste of time" argument. But, that is not a valid argument against the truth....

Movies of aircraft flying in close proximity to clouds further supports this idea that the plane's lift is a product of interaction with the air alone. A plane has to be darn close to a cloud to produce a disturbance in the formation. By the time it's two to three wingspans away there's no effect. So again it shows that the airmass itself is able to deal with the needs for flying the plane without the need to connect with the earth.

Now that we've revised our theory to reflect the fact that the downward "push" exerted by the plane on the earth need not involve downward momentum (i.e. it's no problem if the downward momentum is damped out in just a few wingspans), your cloud observation is not a problem. Again, imagine if we are flying in honey-- due to shear forces, the honey would transmit the downward "push" of the wing all the way to the ground (or all the way to the edge of the container) even if the actual downward momentum of the honey dampened out to near zero in just a very short distance.

For the same reason your space craft example would not produce a downwash velecity vector which traveled for thousands of miles or light years.

If the spacecraft is flying through a very dispersed (non-viscous) gas cloud, whose molecules are acting much like bb's in a vacuum, then I say that if the wing of the space craft puts the molecules (bb's) into motion, they will indeed travel for thousands of miles or light years